1. ## Finding derivatives

y= (x-2)/(x^2-x+1)

Find y and y

for the problem i get to
y=(-x^2+4x-1)/(x^2-x+1)^2 by u/v method but the answer of this derivative problem is (Given in the book)
-3(x^2-3x+1)/(x^2-x+1)^3
How can i Get to this form ?
Thanks

2. $y=\frac{x-2}{x^2-x+1}$

I used the quotient rule (I think you did too):

$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u \frac{dv}{dx}}{v^2}$

where u is the numerator and v is the denominator. In this case:

$u=x-2$ and $v=x^2-x+1$.

So I did it like this:

$\frac{dy}{dx}=\frac{(x^2-x+1)(1)-(x-2)(2x-1)}{(x^2-x+1)^2}$

$\frac{dy}{dx}=\frac{x^2-x+1-2x^2+5x-2}{(x^2-x+1)^2}$

$\frac{dy}{dx}=\frac{-x^2+4x-1}{(x^2-x+1)^2}$

So you're right!

I think it's a typo or just this in another form.

3. Originally Posted by Showcase_22

I think it's a typo or just this in another form.
Yeah! but i have to come in that form somehow and i want to know the way.Besides, i have to go for second derivative of this function too

4. If you put x=1 into both the equations you get two different answers so they're not the same.

The only way to get $(x^2-x+1)^3$ as the denominator is to differentiate again. I thought this would be f''(x).
However, I began to differentiate again (using the quotient rule, it's similar to before) and I got a cubic appearing on top (hence this can't be f''(x)).

Try and differentiate f'(x) again and see what you get. I'm pretty sure i'm doing it correctly.

5. Originally Posted by Showcase_22
If you put x=1 into both the equations you get two different answers so they're not the same.

The only way to get $(x^2-x+1)^3$ as the denominator is to differentiate again. I thought this would be f''(x).
However, I began to differentiate again (using the quotient rule, it's similar to before) and I got a cubic appearing on top (hence this can't be f''(x)).

Try and differentiate f'(x) again and see what you get. I'm pretty sure i'm doing it correctly.
Answer of f(x) seems to be
6x(2x^2-8x+5)/(x^2-x+1)^4 from the book

6. Well I think we can safely say the answer in the back of the book for f'(x) is a typo.

The good news is that you're doing it correctly!

7. Originally Posted by Showcase_22
Well I think we can safely say the answer in the back of the book for f'(x) is a typo.
Actually i m not sure about it. So, anybody here can try this one and make a comment.This answer is declared correct by my instructor.By the way it is the problem from anton's book.